Does ζ(3)\zeta(3) have a connection with π\pi?

The problem

Can be ζ(3) written as απβ, where (α,βC), β0 and α doesn’t depend of π (like 2, for example)?

Details

Several ζ values are connected with π, like:

ζ(2)=π2/6

ζ(4)=π4/90

ζ(6)=π6/945

and so on for all even numbers.

See this mathworld link to more details: Riemann Zeta Function

So the question is, could ζ(3) be written as:

ζ(3)=απβ
α,βC
β0
α not dependent of π

See α not essencially belongs Q and α,β could be real numbers too.

When I wrote α is not dependent of π it’s a strange and a hard thing to be defined, but maybe α can be written using e or γ or 2 or some other constant.

Edit:

Maybe this still a open question. If

2k=0(1)kB2k B22k+2(2k)! (22k+2)!

in 42k=0(1)kB2k B22k+2(2k)! (22k+2)!π3 be of the form δπ3 with δ not dependent of π

and 2k1k3e2πk1 not dependent of π too, this question still hard and open.

Edit 2:

I discovered a result, but later I’ve seen that this is something already known, either way, it is an interesting one to have it here.

ζ(3)=4π2ζ(2)

but, if ζ(2) is of the form απ2, with α not dependent of π, then this still remains as a hard and an open question.

I have a conjecture that ζ(2) will not cancel the π2 term, but since I wasn’t able to prove it and I can’t use it here.

We can express ζ of odd numbers with ζ in a easy way, with a “closed” form like this one.

Answer

The question is whether or not ζ(3) is connected with π. The answer is yes. Moreover, ζ(3)=βπα for some complex α,β. Take α=3 and β=0.0387682…. It is not known whether β=0.0387682… is algebraic or transcendental. It is known, however, that ζ(3) is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If α,β>0 such that αβ=π2, then for each non-negative integer n,
αn(ζ(2n+1)2+k1k2n1e2kα1)=(β)n(ζ(2n+1)2+k1k2n1e2kβ1)
22nn+1k=0(1)kB2k B2n2k+2(2k)! (2n2k+2)!αnk+1βk.
where Bn is the nth-Bernoulli number.

For odd positive integer n, we take α=β=π,
ζ(2n+1)=22n(n+1k=0(1)kB2k B2n2k+2(2k)! (2n2k+2)!)π2n+12k1k2n1e2πk1.

In particular, for n=1,
ζ(3)=4(2k=0(1)kB2k B22k+2(2k)! (22k+2)!)π32k1k3e2πk1.

Observe that the coefficient of π3 is rational, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that ζ(3)π3 is transcendental.

Update: Recently, Takaaki Musha claims to have proved that ζ(2n+1)(2π)2n+1 is irrational for positive n1. However, some objection has since been raised (read comments below).

Attribution
Source : Link , Question Author : GarouDan , Answer Author : user02138

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